
Re: [HM] Bombelli: Cube root of Complex Nos.
Posted:
Feb 14, 2001 1:53 PM


Oops, so sorry! cos(arctan 11/2) would not be hard to find. What I meant to write was cos[(arctan 11/2)/3]. And, indeed, as one quickly realizes, this leads to the angletripling formula (4 cos^3 x  3 cos x = cos 3x). And, indeed, circularly, one basically ends up with one's original cubic!
Bonnie Shulman

